Answer to Question #222708 in Differential Equations for andrew

Question

Find the general solution

1) x³y'''-4x²y''+8xy'-8y=4lnx

Answer 1

Let us solve the differential equation $x^3y'''-4x^2y''+8xy'-8y=4\ln x.$ Let us use the transformation $x=e^t.$ Then $y'_x=y'_t\cdot t'_x=y'_t\cdot \frac{1}{x'_t}=y'_t\cdot e^{-t},$ $y''_{x^2}=(y''_{t^2}\cdot e^{-t}-y'_t\cdot e^{-t})e^{-t}=(y''_{t^2}-y'_t)e^{-2t},$ $y'''_{x^3}=(y'''_{t^3}-3y''_{t^2}+2y'_t)e^{-3t}.$

It follows that we have the following equation $e^{3t}(y'''_{t^3}-3y''_{t^2}+2y'_t)e^{-3t}-4e^{2t}(y''_{t^2}-y'_t)e^{-2t}+8e^ty'_te^{-t}-8y=4t,$ which is equivalent to the equation $y'''_{t^3}-7y''_{t^2}+14y'_t-8y=4t.$ The characteristic equation $k^3-7k^2+14k-8=0$ is equivalent to $(k-1)(k^2-6k+8)=0,$ and to $(k-1)(k-2)(k-4)=0,$ and hence has the roots $k_1=1, k_2=2$ and $k_3=4.$ The general solution of the differential equation $y'''_{t^3}-7y''_{t^2}+14y'_t-8y=4t$ is $y(t)=C_1e^t+C_2e^{2t}+C_3e^{4t}+y_p(t),$ where $y_p(t)=at+b.\\$ It follows that $y_p'(t)=a,y_p''(t)=0,y_p'''(t)=0.$ We have that $14a-8(at+b)=4t.$ Then $-8a=4, 14a-8b=0,$ and thus $a=-\frac{1}{2}, b=\frac{7}{4}a=-\frac{7}{8}.$ Therefore, $y(t)=C_1e^t+C_2e^{2t}+C_3e^{4t}-\frac{1}{2}t-\frac{7}{8}.\\$ We conclude that the general solutions of the differential equation $x^3y'''-4x^2y''+8xy'-8y=4\ln x$ is $y(x)=C_1x+C_2x^2+C_3x^4-\frac{1}{2}\ln x-\frac{7}{8}.$